Difference between revisions of "2020 AMC 12A Problems/Problem 15"
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MRENTHUSIASM (talk | contribs) (Described the detailed steps to find the roots of both equations. lopkiloinm should be the primary author and I am secondary. I will use Asymptote for diagram later.) |
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<math>\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}</math> | <math>\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}</math> | ||
− | == Solution == | + | == Solution 1 == |
− | + | We solve each equation separately: | |
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>We solve <math>z^{3}-8=0</math> by De Moivre's Theorem.<p> | ||
+ | Let <math>z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,</math> where <math>r</math> is the magnitude of <math>z</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> <p> | ||
+ | We have <cmath>z^3=r^3\operatorname{cis}(3\theta)=8(1+0i),</cmath> from which | ||
+ | <ul style="list-style-type:square;"> | ||
+ | <li><math>r^3=8,</math> so <math>r=2.</math></li><p> | ||
+ | <li><math>\begin{cases} | ||
+ | \begin{aligned} | ||
+ | \cos(3\theta) &= 1 \\ | ||
+ | \sin(3\theta) &= 0 | ||
+ | \end{aligned}, | ||
+ | \end{cases}</math> so <math>3\theta=0,2\pi,4\pi,</math> or <math>\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}.</math> </li><p> | ||
+ | </ul> | ||
+ | The set of solutions to <math>z^{3}-8=0</math> is <math>\boldsymbol{A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}}.</math> In the complex plane, they are the vertices of an equilateral triangle whose circumcircle has center <math>0</math> and radius <math>2.</math></li><p> | ||
+ | <li>We solve <math>z^{3}-8z^{2}-8z+64=0</math> by factoring by grouping.</li><p> | ||
+ | We have | ||
+ | <cmath>\begin{align*} | ||
+ | z^2(z-8)-8(z-8)&=0 \\ | ||
+ | \bigl(z^2-8\bigr)\bigl(z-8\bigr)&=0. | ||
+ | \end{align*}</cmath> | ||
+ | The set of solutions to <math>z^{3}-8z^{2}-8z+64=0</math> is <math>\boldsymbol{B=\left\{2\sqrt{2}i,-2\sqrt{2}i,8\right\}}.</math> | ||
+ | </ol> | ||
+ | In the graph below, the greatest distance between a point of <math>A</math> and a point of <math>B</math> is the distance between <math>-1\pm\sqrt{3}i</math> to <math>8,</math> as shown in the dashed line segments. By the Distance Formula, the answer is <cmath>\sqrt{(8-(-1))^2+\left(\pm\sqrt{3}-0\right)^2}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}}.</cmath> | ||
+ | <b>DIAGRAM NEEDED.</b> | ||
− | + | ~lopkiloinm ~MRENTHUSIASM | |
− | |||
− | |||
− | |||
− | ~lopkiloinm | ||
==Remark== | ==Remark== |
Revision as of 04:21, 11 September 2021
Contents
Problem
In the complex plane, let be the set of solutions to
and let
be the set of solutions to
What is the greatest distance between a point of
and a point of
Solution 1
We solve each equation separately:
- We solve
by De Moivre's Theorem.
Let
where
is the magnitude of
such that
and
is the argument of
such that
We have
from which
so
so
or
is
In the complex plane, they are the vertices of an equilateral triangle whose circumcircle has center
and radius
- We solve
by factoring by grouping.
We have
The set of solutions to
is
In the graph below, the greatest distance between a point of and a point of
is the distance between
to
as shown in the dashed line segments. By the Distance Formula, the answer is
DIAGRAM NEEDED.
~lopkiloinm ~MRENTHUSIASM
Remark
In the graph below, the solutions to are shown in red, and the solutions to
are shown in blue. The greatest distance between one red point and one blue point is shown in a black dashed line segment.
Graph in Desmos: https://www.desmos.com/calculator/uylcxkffak
~MRENTHUSIASM
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.